The Basic Rules of Counting

# 1 and n k n 0 1 n nk math 30530 fall 2012 counting

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Unformatted text preview: then those without both John and Pat 8 7 8 6.5.4 + 5.4 + 6.5.4 = 10, 780 5 4 6 How many anagrams of MISSISSPPI? 11!/(4!4!2!1!) = 34, 650 Math 30530 (Fall 2012) Counting September 13, 2013 6 / 12 Some properties of binomial coefﬁcients n n = 1 (justiﬁes 0! = 1) and Math 30530 (Fall 2012) n 0 =1 Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients n n = 1 (justiﬁes 0! = 1) and n k = n 0 =1 n n−k Math 30530 (Fall 2012) Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients n n = 1 (justiﬁes 0! = 1) and n k = n 0 =1 n n−k (x + y )n = Math 30530 (Fall 2012) n n k =0 k x k y n−k (Binomial Theorem) Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients n n = 1 (justiﬁes 0! = 1) and n k = n 0 =1 n n−k (x + y )n = Special case: Math 30530 (Fall 2012) n n k =0 k x k y n−k (Binomial Theorem) n n k =0 k = 2n Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients n n = 1 (justiﬁes 0! = 1) and n k = n 0 =1 n n−k (x + y )n = Special case: n n k =0 k x k y n−k (Binomial Theorem) n n k =0 k = 2n If I repeat the same trial n times, independently, and each time I have probability p of success, then the probability that I have n exactly k successes is k pk (1 − p)n−k Math 30530 (Fall 2012) Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients n n = 1 (justiﬁes 0! = 1) and n k = n 0 =1 n n−k (x + y )n = Special case: n n k =0 k x k y n−k (Binomial Theorem) n n k =0 k = 2n If I repeat the same trial n times, independently, and each time I have probability p of success, then the probability that I have n exactly k successes is k pk (1 − p)n−k Many identities, e.g. Committee/Chair: n Math 30530 (Fall 2012) n−1 k −1 =k n k Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients n n = 1 (justiﬁes 0! = 1) and n k = n 0 =1 n n−k n n k =0 k (x + y )n = Special case: x k y n−k (Binomial Theorem) n n k =0 k = 2n If I repeat the same trial n times, independently, and each time I have probability p of success, then the probability that I have n exactly k successes is k pk (1 − p)n−k Many identities, e.g. n−1 k −1 = n −1 n −1 k −1 + k Committee/Chair: n Pascal: Math 30530 (Fall 2012) n k = k n k...
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## This note was uploaded on 01/31/2014 for the course MATH 30530 taught by Professor Hind,r during the Fall '08 term at Notre Dame.

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